First of all, we are dealing with unitary representations, so that the $T^a$s are always self-adjoint and the representations have the form
$$U(v) = e^{i \sum_{a=1}^Nv^a T_a}$$
with $v \in \mathbb R^N$. When you say that $U$ is real you just mean that the representation is made of the very real, unitary, $n\times n$ matrices $U$. This way, the condition $(T_a)^* = -T_a$ is equivalent to the reality (in the proper sense) requirement.
Let us come to your pair of questions.
(1). You are right on your point:
PROPOSITION. A unitary finite dimensional representation is complex (i.e. it is neither real nor pseudoreal) if and only if at least one self-adjoint generator $T_a$ has an eigenvalue $\lambda$ such that $-\lambda$ is not an eigenvalue.
PROOF
Suppose that $$(T_a)^* = -V T_a V^{-1}\tag{1}$$ for some unitary matrix $V$ and every $a=1,2,3,\ldots, N$. Since we also know that $T_a$ is self-adjoint, there is an orthogonal basis of eigenvectors $u_j^{(a)}\neq 0$, $j=1,\ldots, n$ and the eigenvalues $\lambda_j^{(a)}$ are real. Therefore:
$$T_au_j^{(a)}= \lambda_j^{(a)} u_j^{(a)}\:.$$
Taking the complex conjugation and using (1)
$$VT_aV^{-1}u_j^{(a)*}= -\lambda_j^{(a)} u_j^{(a)*}$$
so that $V^{-1}u_j^{(a)*}\neq 0$ is an eigenvector of $T_a$ with eigenvalue $-\lambda_j$. We conclude that $\lambda$ is an eigenvalue if and only if $-\lambda$ is (consequently, if the dimension of the space is odd, $0$ must necessarily be an eigenvalue as well).
Soppose,vice versa, that for the self-adjoint matrix $T^a$ its (real) eigenvalues satisfy the constraint that $\lambda$ is an eigenvalue if and only if $-\lambda$ is. As $T^a$ is self adjoint, there is a unitary matrix such that:
$$T_a = U diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2}) U^{-1}$$
when $n$ is even, otherwise there is a further vanishing last element on the diagonal.
Thus
$$T^*_a = U^* diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2}) U^{-1*}$$
Notice that $U^*$ is unitary if $U$ is such.
Let us indicate by $e_1,e_2, \ldots, e_n$ the orthonormal basis of eigenvectors of $T^a$ where the matrix takes the above diagonal form.
If $W$ is the (real) unitary matrix which swaps $e_1$ with $e_2$, $e_3$ with $e_4$ and so on (leaving $e_n$ fixed if $n$ is odd), we have that
$$W diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2}) W^{-1} = - diag(\lambda_1, -\lambda_1, \lambda_2, -\lambda_2,..., \lambda_{n/2},-\lambda_{n/2})$$ and thus
$$T^*_a = U^*W^{-1}(- UT_aU^{-1}) WU^{-1*} = -S T_a S^{-1}$$
with $S= U^*W^{-1}U$, which is unitary because composition of unitary matrices.
We can conclude that, as you claim, a way to show that a representation is complex (i.e. it is not real) is to show that at least one generator matrix $T_a$ has (non-vanishing) eigenvalues that do not come in plus-minus pairs.
QED
(2). In view of (1), if the list of eigenvalues you presented is correct, the considered representation is obviously complex.
